Problem: The graphs of $y=|x|$ and $y=-x^2-3x-2$ are drawn.  For every $x$, a vertical segment connecting these two graphs can be drawn as well.  Find the smallest possible length of one of these vertical segments.
Answer: The function $|x|$ is difficult to deal with directly.  Instead we work by cases: $x\geq0$ and $x<0$.

If $x\geq0$ then $|x|=x$, and we can find the difference by subtracting  \[x-(-x^2-3x-2)=x^2+4x+2=(x+2)^2-2.\]This function is always increasing as $x$ varies over the nonnegative numbers, so this is minimized at $x=0$. The minimum value on $x\geq0$ is  \[(0 + 2)^2 - 2 = 2.\]If $x<0$ then $|x|=-x$ and we can find the difference by subtracting: \[(-x)-(-x^2-3x-2)=x^2+2x+2=(x+1)^2+1.\]This quadratic is minimized at $x=-1$, and the minimum value is  \[(-1+1)^2+1=1.\]Since the minimum value on negative numbers is less than the minimum value on the nonnegative numbers, the minimum value for the difference is $\boxed{1}$.